2021 AMC 12A Fall 第 16 题

先试着解答 2021 AMC 12A Fall 第 16 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2021 AMC 12A Fall 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

16.

一个组织有 3030 名员工,其中 2020 人使用 A 品牌电脑,另外 1010 人使用 B 品牌电脑。出于安全原因,电脑只能彼此相连,且只能用电缆连接。电缆只能把一台 A 品牌电脑和一台 B 品牌电脑相连。如果两名员工的电脑直接由电缆相连,或可以通过一系列相连的电脑转发消息,则他们可以相互通信。最初没有任何电脑与其他电脑相连。一名技术员任意选择一台每种品牌的电脑,并在它们之间安装电缆,前提是这对电脑之间还没有电缆。技术员在每名员工都能相互通信时停止。最多可能使用多少根电缆?
An organization has 3030 employees, 2020 of whom have a brand A computer while the other 1010 have a brand B computer. For security, the computers can only be connected to each other and only by cables. The cables can only connect a brand A computer to a brand B computer. Employees can communicate with each other if their computers are directly connected by a cable or by relaying messages through a series of connected computers. Initially, no computer is connected to any other. A technician arbitrarily selects one computer of each brand and installs a cable between them, provided there is not already a cable between that pair. The technician stops once every employee can communicate with each other. What is the maximum possible number of cables used?
190190
191191
192192
195195
196196
答案:B
知识点:图论极端原理
小提示:
为了尽量延迟连通,尽可能久地保持图不连通,然后最后一根电缆把所有部分连接起来
To delay connectivity, keep the graph disconnected as long as possible, then one final cable joins everything
大提示:
在仍不连通的情况下,最多边数来自 1919 台 A 品牌和 1010 台 B 品牌形成的完全二分图,另留一台 A 品牌电脑孤立
The most edges while still disconnected is a complete bipartite graph on 1919 brand A and 1010 brand B, leaving one brand A computer isolated
解答:
技术员不断加电缆,直到图变为连通。为了最大化电缆数,要让网络尽可能久地保持不连通:留下一台 A 品牌电脑孤立,并把其余 1919 台 A 品牌电脑与全部 1010 台 B 品牌电脑完全相连。
这样在仍不连通时使用了 19⋅10=19019 \cdot 10 = 190 根电缆。下一根电缆连接最后一台 A 品牌电脑,使所有人连通,总数为 190+1=191190 + 1 = 191。
所以正确答案是 B。
The technician keeps adding cables until the graph becomes connected. To maximize the count, keep the network disconnected for as long as possible: leave a single brand A computer isolated and fully connect the remaining 1919 brand A computers to all 1010 brand B computers.
That uses 19⋅10=19019 \cdot 10 = 190 cables while still disconnected. The next cable connects the last brand A computer, joining everyone, for a total of 190+1=191.190 + 1 = 191.
Thus, the correct answer is B.
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